Exponential revision question I can't work out?

1. Nov 6, 2014

Ramjam

< Mentor Note -- Thread moved from the Calculus forum to HH/Calculus >

Hey everyone, I've started my new course and have been given revision and tutorial questions to do at home, this question has been bugging me for days now! I need to make "t" the subject to allow me to progress into the solution further.

Make t the subject in " i=12.5(1-e^-t/cr) "
I know that i need to use natural logs (ln) however i cannot seem to transpose the equation correctly. Could you please help me transpose this.

Thanks
Ramjam

p.s forgot to add my working so far, DUH!
ive transposed it to this form so far
i/12.5 = 1 - e ^-t/cr

this is where the logs come in and i have trouble, im not sure what to do next?!?

Last edited by a moderator: Nov 6, 2014
2. Nov 6, 2014

pasmith

Before taking logs, you must make $e^{-t/cr}$ the subject.

3. Nov 6, 2014

Staff: Mentor

This is very unclear.
The quantity in parentheses could be any of the following:
$1 - \frac{e^{-t}}{cr}$
$1 - \frac{e^{-t}}{c}r$
$1 - e^{-t/c * r}$
No, logs don't come in yet. You still have that 1 on the right side.

4. Nov 6, 2014

Ramjam

How do I go about removing the one on the right hand side, is it correct for me to move it to the left hand side, leaving e^-t/cr on the right?

5. Nov 6, 2014

Ray Vickson

He might even mean $\frac{1-e^{-t}}{cr}$. For some unknown reason, he refuses to deal with your question.

6. Nov 6, 2014

Ramjam

My lecturer hinted that the 1 should be moved leaving e^t/cr on the right, however unsure how to transpose it to that

7. Nov 6, 2014

vela

Staff Emeritus
I moved this to the pre-calc forum since the issue is apparently about basic algebra.

8. Nov 6, 2014

9. Nov 6, 2014

Ramjam

1-e-t/cr sorry my phone cut your post so I didn't understand what you meant

10. Nov 6, 2014

Staff: Mentor

You have
What can you do to both sides of this equation so that the 1 on the right side is gone? Possible choices are to add the same number toboth sides, subtract the same number from both sides, multiply both sides by the same nonzero number, or divide both sides by the same nonzero number.

11. Nov 6, 2014

Ramjam

If I was to add 1 to both sides e.g i/12.5 + 1 = 1-1e-t/cr is this acceptable? As then the one can be removed from the right ?

12. Nov 6, 2014

Staff: Mentor

No, plus it's wrong.
i/12.5 = 1 - e-t/cr

If you add 1 to both sides, you get
i/12.5 + 1 = 1 - e-t/cr + 1 = 2 - e-t/cr
So that's no help at all.

Note that 1e-t/cr is exactly the same as e-t/cr, so there's no point in writing a coefficient of 1 here.

Nope.

13. Nov 6, 2014

Ramjam

Could you possibly take me through step by step how to remove 1 from the right leaving e-t/cr so that I can see how it's done, as confusing my self now

Ramjam

14. Nov 6, 2014

Staff: Mentor

There is no "step by step" here - this is not a complicated process. Earlier in this thread you said,
You added 1 to the left side, but you didn't add 1 to the right side. Whatever you do to one side of the equation, you have to do the same to the other side.

If you have i/12.5 = 1 - e-t/cr, what can you add to the 1 on the right side to get rid of it? In other words, what + 1 = 0?

15. Nov 6, 2014

Ramjam

Okay so adding -1 to both sides i/12.5 -1 = 1-e-t/cr -1
which when becomes i/12.5 -1 = -e-t/cr

isnt the minus infront the the "e" become an issue? as you can log "-e"

Thankyou for your help on the matter

16. Nov 6, 2014

Staff: Mentor

You can still multiply both sides by the same factor...
(and you should try those basic steps yourself)

17. Nov 6, 2014

Staff: Mentor

Yes.
So what can you do to get rid of the minus sign? Remember to do the same thing to both sides of the equation.

18. Nov 6, 2014

Ramjam

Umm..... Im not actually to sure? :/

19. Nov 6, 2014

Staff: Mentor

Then try different things. At least one of them will work.

20. Nov 6, 2014

Ramjam

Well ive been trying things for the past half hour with no luck, im not seeing how i can multiply both sides to remove the minus from the "e"